A proton enters a uniform magnetic field that is at a right angle to its velocit

Question

A proton enters a uniform magnetic field that is at a right angle to its velocity. The field strength is 0.90 and the proton follows a circular path with a radius of 4.8 .


What is the magnitude of its linear momentum?
Express your answer using two significant figures.
=




Part B
What is its kinetic energy?
Express your answer using two significant figures.
=




Part C
If its speed were doubled, what would then be the radius?
doubling the speed would double the radius
doubling the speed would increase the radius by a factor of 4
doubling the speed would decrease the radius by a factor of 2


Part D
What would be the momentum?
doubling the speed would increase the momentum by a factor of 4
doubling the speed would decrease the momentum by a factor of 4
doubling the speed would double the momentum


Part E
What would be the kinetic energy?
doubling the speed would increase the kinetic energy by a factor of 4
doubling the speed would double the kinetic energy
doubling the speed would decrease the kinetic energy by a factor of 4

Explanation / Answer

B = 0.90 T, r = 4.8 cm, q = 1.6*10-19 C, m = 1.67*10-27 kg

What is the magnitude of its linear momentum?
qvB = mv2/r
p = mv = qBr = 6.9*10-21 kgm/s

Part B
What is its kinetic energy?
K = p2/(2m) = 1.4*10-14 J

Part C
If its speed were doubled, what would then be the radius?

r = mv/(qB), so

doubling the speed would double the radius

Part D
What would be the momentum?

p = mv, so

doubling the speed would double the momentum

Part E
What would be the kinetic energy?

K = mv2/2, so

doubling the speed would increase the kinetic energy by a factor of 4

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